Work and Time Questions for UP POLICE CONSTABLE

Core Concept: Work Rate as a Fraction

Work is always 1 complete job. If person A completes a job in 10 days, their work rate is 1/10 of the job per day. Think of work as a pie: if you eat 1/10 of the pie each day, you finish in 10 days. The faster you work, the larger your fraction per day. This fraction is your 'work rate' — the foundation of all work-time problems.

• Work rate = 1 / (days to complete job alone)
• If A does 1/10 work per day, A finishes in 10 days
• Work rate is always positive and expressed as a fraction or decimal
• Total work done = work rate × number of days worked
• When multiple people work together, ADD their individual rates

Work Left and Partial Completion

Many UP Police questions ask: 'A works for 3 days, then B joins. How long to finish?' Break this into phases. First calculate how much work A completes alone, subtract from total (1), then find combined rate for remaining work. This two-phase approach is critical for multi-person, multi-stage problems.

• Work left = 1 - (work completed in phase 1)
• Phase 1: One person works alone for X days
• Phase 2: Multiple people work together on remaining work
• Always track cumulative work; it must sum to exactly 1
• If work is not 1 (e.g., 'paint 5 walls'), treat 5 walls as your unit = 1 job

Efficiency and Ratio-Based Work

Some questions state: 'A is twice as efficient as B' or 'A:B work in ratio 3:2'. Efficiency is directly proportional to work rate. If A is twice as efficient, A's rate is 2x and B's rate is x. Use ratio logic to split the work or time inversely: if rates are in ratio 3:2, time taken alone is in ratio 2:3 (inverse). This inverse relationship is a common trap.

• Efficiency ratio = Work rate ratio
• If A is 3 times as efficient as B, then Rate_A = 3 × Rate_B
• Time ratio is INVERSE of efficiency ratio: if efficiency is 3:2, time is 2:3
• In ratio problems, use variables: let B's rate = x, then A's rate = 3x (if 3:1 ratio)
• Always verify: higher efficiency = shorter time to complete alone

Work Done by Fraction of Workforce

Questions like 'If 5 men do a job in 12 days, how long for 8 men?' assume all workers are equally efficient. Total work = number of workers × days × work per person per day. This is the 'man-days' concept. If work is constant, men × days = constant. Doubling workers halves the time (inverse proportion).

• Total work (in man-days) = number of workers × number of days
• If work is fixed, workers and days are inversely proportional
• 5 men × 12 days = 60 man-days of work
• Same 60 man-days with 8 men = 60/8 = 7.5 days
• Useful for 'if more workers join' or 'some workers leave' scenarios

Pipes and Cisterns (Special Case)

Pipes filling/emptying tanks follow the same rate logic. Inlet pipe fills at rate 1/x (tank per day), outlet drains at rate 1/y. When both open, net rate = 1/x - 1/y (inlet minus outlet). If net is positive, tank fills; if negative, it empties. This is work-time applied to fluid flow — same math, different context.

• Inlet pipe alone fills in X days: rate = 1/X per day
• Outlet pipe alone empties in Y days: rate = -1/Y per day (negative because it removes)
• Both open together: net rate = 1/X - 1/Y
• If net rate is negative, tank will never fill (drain is faster)
• Time to fill = 1 / net rate (if net is positive)

• Forgetting to INVERT the ratio: if efficiency is 3:2, time is 2:3, not 3:2. Aspirants often use the same ratio for both, leading to wrong answers.
• Adding rates incorrectly: always find a common denominator. 1/12 + 1/15 is NOT 2/27. Use LCM: 1/12 + 1/15 = 5/60 + 4/60 = 9/60.
• Confusing 'work left' with 'total work': if A does 1/4 of job, work left is 3/4, not 1/4. Subtract from 1, not from the original time.
• Ignoring the 'both open' condition in pipes: students forget to subtract outlet rate, treating it as addition. Outlet is negative.
• Assuming all workers are equally efficient without being told: some questions specify different efficiencies. Read carefully — 'A is twice as fast' vs. 'A and B work equally'.

• PIE SLICING: Think of work as a pie. If you eat 1/10 per day, you finish in 10 days. Faster eating = bigger slice per day = fewer days total.
• INVERSE INVERSION: Efficiency and time are inverses. 3x efficiency = 1/3 the time. Write it down: if A is 3 times faster, A takes 1/3 the days.
• RATE + RATE = COMBINED RATE: When people work together, ADD their rates. When pipes are both open, ADD inlet, SUBTRACT outlet.
• MAN-DAYS CONSTANT: 5 men × 12 days = 60 man-days. If you change men, divide 60 by new men count to get new days. Inverse proportion.

• UP Police typically asks 2-3 work-time questions per paper. Expect one straightforward 'A and B together' problem and one multi-stage (A works alone, then B joins) problem.
• Time pressure is real: practice converting fractions quickly. LCM of 12 and 15 should be instant (60). Use shortcuts: 1/12 + 1/15 = (5+4)/60 = 9/60 = 3/20.
• Pipes and cisterns appear 1-2 times per paper. Remember: inlet is positive, outlet is negative. If both open and outlet is faster, tank empties (negative net rate).
• Efficiency-ratio problems are common in recent papers. If question says 'A is 3 times as efficient', immediately set Rate_A = 3x, Rate_B = x. Solve for x using the 'together' condition.
• Always double-check by substitution: if A takes 6.25 days alone and B takes 25 days, combined should be less than 6.25. Verify: 1/6.25 + 1/25 = 0.16 + 0.04 = 0.2 = 1/5 days. Yes, 5 days is less than 6.25. Logic check passes.